Find the vector equation of the plane that contains the lines `vecr = (hat"i" + hat"j") + λ (hat"i" + 2hat"j" - hat"k")` and the point (–1, 3, –4). Also, find the length of the perpendicular drawn from the point (2, 1, 4) to the plane thus obtained.

#### Solution

Let the vector equation of the required plane be `vec"r" . vec"n" = d`

The plane contains the line `vec"r" = hat"i" + hat"j" + λ (hat"i" + 2hat"j" - hat"k")`

Since the plane passes through point A and B. So `vec"n"` will be parallel to vector `vec"AB" xx (hat"i" + 2hat"j" - hat"k")`

`vec"AB" = vec"OB" -vec"OA"`

= `(-hat"i" + 3hat"j" - 4hat"k") (hat"i" + hat"j")`

= `-2hat"i" + 2hat"j" - 4hat"k"`

`vec"AB" xx (hat"i" + 2hat"j" - hat"k") = |(hat"i",hat"j",hat"k"),(1,2,-1),(-2,2, -4)|`

= `hat"i" (-8 + 2) - hat"j" (-4-2) + hat"k" (2+4)`

= `-6hat"i" + 6hat"j" + 6hat"k"`

which is a normal vector to the plane.

So the equation of plane will be `vec"r" . (-6hat"i" + 6hat"j" + 6hat"k") = d`

∴ it passes through (1, 1, 0) so `(hat"i" + hat"j"). (-6hat"i" + 6hat"j" + 6hat"k") = d or, d = 0`

equation of plane is `vecr . (-6hat"i" + 6hat"j" + 6hat"k") = 0`

`vecr (hat"i" - hat"j" -hat"k") = 0`

in Cartesian plane,

`(xhat"i" + yhat"j" + zhat"k") . (hat"i" - hat"j" -hat"k") = 0`

x -y - z = 0

So, the perpendicular distance of the plane from the point (2, 1, 4) is ` = |(2 -1-4)/sqrt(1^2 + (-1)^2 + (-1)^2)| = |(-3)/sqrt(3)| = sqrt(3) "unit"`.